CHAPTER 13Kinetics of Particles:
Energy and
Momentum Methods
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Chapter OutLine
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1. Introduction
2. Work of a Force
3. Principle of Work & Energy
4. Applications of the Principle
of Work & Energy
5. Power and Efficiency
6. Potential Energy
7. Conservative Forces
8. Conservation of Energy
10. Motion Under a Conservative
Central Force
11. Principle of Impulse and
Momentum
12. Impulsive Motion
13. Impact
14. Direct Central Impact
15. Oblique Central Impact
Introduction
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• Chapter 12 deals with motion of particles
through the equation of motion,
amF
Introduction
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Chapter 13: Two additional methods:
Method of work & energy: relates force,
mass, velocity and displacement
Method of impulse &momentum: relates
force, mass, velocity, and time
Work of a Force
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= particle displacementrd
Work of force is
dzFdyFdxF
dsFrdFdU
zyx
cos
Work is a scalar quantity
Work = area under Ft
vs. s curve
Work of a Force
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Work of a force during
a finite displacement,
2
1
2
1
2
1
2
1
cos21
A
A
zyx
s
s
t
s
s
A
A
dzFdyFdxFdsF
dsFrdFU
Work of a Force
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Work of a constant force in rectilinear motion,
xFU cos21
Work of the force of gravity,
yWyyW
dyWU
dyW
dzFdyFdxFdU
y
y
zyx
12
21
2
1
Work of a Force - gravity
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Work of weight is positive when y < 0, when
weight moves down
Work of a Force - spring
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Magnitude of force exerted by a spring is
proportional to deflection,
kxF
Work of force exerted by spring,
222
1212
121
2
1
kxkxdxkxU
dxkxdxFdU
x
x
Work of a Force - spring
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Work of force exerted by
spring is positive when x2 <
x1, when spring is returning
to its undeformed position
Work of force exerted by
spring = negative of area
under F vs. x curve,
xFFU 2121
21
Work of a Force - gravitational
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)11
(12
221
2
2
1
rrGMm
drr
MmGU
drr
MmGFdrdU
r
r
Forces which do not do work (ds = 0 or cos 0:
Weight of a body when its center of gravity
moves horizontally
Reaction at a roller moving along its track
Reaction at frictionless surface when body in
contact moves along surface
Reaction at frictionless pin supporting rotating
body
Work of a Force
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Particle Kinetic Energy: Principle
of Work & Energy
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dvmvdsF
ds
dvmv
dt
ds
ds
dvm
dt
dvmmaF
t
tt
A particle of mass m acted upon by force F
2
21
1221
2
1212
221
2
1
2
1
mvTTTU
mvmv
dvvmdsF
v
v
s
s
t
Applications of the Principle of
Work and Energy
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Determine velocity of bob at A2
glv
vg
WWl
TUT
2
2
10
2
22
2211
Velocity found without determining expression for
acceleration and integrating
All quantities are scalars and can be added directly
Forces which do no work are eliminated from
problem
Applications of the Principle of
Work and Energy
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As the bob passes through A2
Wl
gl
g
WWP
l
v
g
WWP
amF nn
32
22
glv 22
Power and Efficiency
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Power = rate at which work is done
vFdt
rdF
dt
dUPower
Units of power:
W746s
lbft550 hp 1or
s
mN 1
s
J1 (watt) W 1
inputpower
outputpower
input work
koutput worefficiency
Sample Problem 13.1
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An automobile weighing 4000 lb is driven down a 5o
incline at a speed of 60 mph when the brakes are
applied causing a constant total breaking force of
1500 lb.
Determine the distance traveled by the automobile
as it comes to a stop.
Sample Problem 13.2
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Two blocks are joined by an inextensible cable as
shown. If the system is released from rest, determine
the velocity of block A after it has moved 2 m.
Assume that the coefficient of friction between block
A and the plane is mk = 0.25 and that the pulley is
weightless and frictionless.
Sample Problem 13.3
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A spring is used to stop a 60 kg package which is sliding on a
horizontal surface. The spring has a constant k = 20 kN/m and
is held by cables so that it is initially compressed 120 mm. The
package has a velocity of 2.5 m/s in the position shown and the
max deflection of the spring is 40 mm. Determine
a. Coefficient of kinetic friction between package & surface
b. Velocity of package as it passes again through the position
shown.
Sample Problem 13.4
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A 2000 lb car starts from rest at point 1 and moves
without friction down the track shown. Determine:
a. Force exerted by the track on the car at point 2
b. Min safe value of radius of curvature at point 3
Sample Problem 13.5
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The dumbwaiter D and its load have a
combined weight of 600 lb, while the
counterweight C weighs 800 lb.
Determine the power delivered by the
electric motor M when the dumbwaiter
a) is moving up at a constant speed of 8
ft/s
b) has an instantaneous velocity of 8
ft/s and an acceleration of 2.5 ft/s2,
both directed upwards.
D
Bonus
Test max speed of a car traveling along
straight line such that to assure stability
of the car using ADAMS software
Points: 2%
Due: Monday 12/3/2018
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Assignment 13-1
1, 8, 15, 22, 28, 43, 50
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Potential EnergyFriday, February 23, 2018
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2121 yWyWU
Work of force of gravity ,W
Work is independent of path
followed; depends only on
initial & final values of Wy
WyVg
2121 gg VVU
Choice of datum is arbitrary
Potential Energy
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For a space vehicle, variation of force of gravity
with distance from center of earth should be
considered.
Work of a gravitational force,
1221
r
GMm
r
GMmU
r
WR
r
GMmVg
2
Potential EnergyFriday, February 23, 2018
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Work of force exerted by a
spring depends only on initial
& final deflections of spring,
2
2212
121
21 xkxkU
2121
2
21
ee
e
VVU
kxV
Conservative Forces
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Work of force is independent of path
followed by its point of application.
Conservation of
Energy
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Work of a conservative
force,2121 VVU
Concept of work & energy,
1221 TTU
constant
2211
VTE
VTVT
WVT
WVT
11
11 0
WVT
VWgg
WmvT
22
2222
12 02
2
1
Conservation of Energy
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When a particle moves under the action of
conservative forces, total mechanical energy is
constant.
Friction forces are not conservative. Total
mechanical energy of a system involving friction
decreases.
Mechanical energy is dissipated by friction into
thermal energy. Total energy is constant.
Motion Under a Conservative Central Force
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Conservation of angular momentum
Conservation of energy
sinsin 000 rmvmvr
r
GMmmv
r
GMmmv
VTVT
221
0
202
1
00
Sample Problem 13.6
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A 20 lb collar slides without
friction along a vertical rod as
shown. The spring attached to
the collar has an undeflected
length of 4 in. and a constant
of 3 lb/in.
If collar is released from rest
at position 1, determine its
velocity after it has moved 6
in to position 2.
Sample Problem 13.7
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The 0.5 lb pellet is pushed against the spring and
released from rest at A. Neglecting friction,
determine the smallest deflection of the spring for
which the pellet will travel around the loop and
remain in contact with the loop at all times.
Assignment 13-2
55, 61, 67, 76, 95
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221121 , of impulse 2
1
vmvmFdtF
t
t
ImpImp
12
2
1
, vmvmdtFvmddtF
t
t
Principle of Impulse & Momentum
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From Newton’s 2nd law,
vmvmdt
dF
linear momentum
Mohammad Suliman Abuhaiba, Ph.D., PE
Principle of Impulse & Momentum
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Final momentum of particle = initial momentum +impulse of force during the time interval
Units for the impulse of a force
smkgssmkgsN 2
Mohammad Suliman Abuhaiba, Ph.D., PE
Impulsive Motion
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Impulsive force: Force acting on a particle during
a very short time interval that is large enough to
cause a significant change in momentum
When impulsive forces act on a particle,
21 vmtFvm
Impulsive Motion
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21 vmtFvm
When a baseball is struck by a bat, contact occurs
over a short time interval but force is large
enough to change sense of ball motion.
Non-impulsive forces: forces for which impulse
is small and therefore, may be neglected.
Mohammad Suliman Abuhaiba, Ph.D., PE
Sample Problem 13.10
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An automobile weighing 4000 lb is driven down a 5o
incline at a speed of 60 mph when the brakes are
applied, causing a constant total braking force of
1500 lb.
Determine the time required for the automobile to
come to a stop.
Sample Problem 13.11
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A 4 oz baseball is pitched
with a velocity of 80 ft/s.
After the ball is hit by the
bat, it has a velocity of
120 ft/s in the direction
shown. If the bat and
ball are in contact for
0.015 s, determine the
average impulsive force
exerted on the ball during
the impact.
Sample Problem 13.12
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A 10 kg package drops from a
chute into a 24 kg cart with a
velocity of 3 m/s. Knowing
that the cart is initially at rest
and can roll freely, determine
a) final velocity of the cart
b) impulse exerted by the cart
on the package
c) fraction of the initial energy
lost in the impact
Assignment 13-3
119, 129, 136, 149, 154
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Impact
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Impact:
• Collision between two bodies
• Occurs during a small time
interval
• During which bodies exert
large forces on each other
Line of Impact: Common
normal to surfaces in contact
during impact
Direct Central Impact
Oblique Central Impact
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Central Impact:
• mass centers of the two
bodies lie on line of impact
• otherwise, it is an eccentric
impact
Direct Impact:
• velocities of the two bodies
are directed along the line
of impact
Direct Central Impact
Oblique Central Impact
Impact
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Oblique Impact: one or both of
the bodies move along a line
other than the line of impact.
Direct Central Impact
Oblique Central Impact
Impact
Direct Central
Impact
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Bodies move in same straight
line, vA > vB
Upon impact bodies undergo a
period of deformation, at end
of which, they are in contact
and moving at a common
velocity.
A period of restitution:
bodies either regain their
original shape or remain
permanently deformed.
Direct Central Impact
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Total momentum of the two
body system is preserved,
BBAABBAA vmvmvmvm
A second relation between
final velocities is required.
Direct Central Impact
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Period of deformation: umPdtvm AAA
Period of restitution: AAA vmRdtum
10
e
uv
vu
Pdt
Rdtnrestitutio of tcoefficien e
A
A
Direct Central Impact
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A similar analysis of particle B yieldsB
B
vu
uv e
Combining the relations leads to:
ABAB vvevv
Perfectly plastic impact, e = 0: vvv AB
vmmvmvm BABBAA
Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.
BAAB vvvv
Oblique Central Impact
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Tangential component of momentum for each
particle is conserved.
tBtBtAtA vvvv
Normal component of total momentum of the two
particles is conserved.
nBBnAAnBBnAA vmvmvmvm
Oblique Central Impact
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Normal components of relative velocities before
and after impact are related by coefficient of
restitution.
nBnAnAnB vvevv
Oblique Central Impact
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Block constrained to move along horizontal surface.
Oblique Central Impact
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Tangential momentum of
ball is conserved. tBtB vv
Total horizontal momentum of block & ball is
conserved: xBBAAxBBAA vmvmvmvm
Normal component of relative velocities of block
and ball are related by coefficient of restitution.
nBnAnAnB vvevv
Problems Involving Energy & Momentum
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Three methods for analysis of kinetics problems:
1. Direct application of Newton’s 2nd law
2. Method of work & energy
3. Method of impulse & momentum
Sample Problem 13.14
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A ball is thrown against a
frictionless, vertical wall.
Immediately before the ball
strikes the wall, its velocity
has a magnitude v and forms
angle of 30o with the
horizontal. Knowing that
e = 0.90, determine the
magnitude and direction of
the velocity of the ball as it
rebounds from the wall.
Sample Problem 13.15
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The magnitude & direction of the velocities of two
identical frictionless balls before they strike each other
are as shown. Assuming e = 0.9, determine the
magnitude & direction of the velocity of each ball after
the impact.
Sample Problem 13.16
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Ball B is hanging from an
inextensible cord. An identical ball
A is released from rest when it is
just touching the cord and acquires
a velocity v0 before striking ball B.
Assuming perfectly elastic impact
(e = 1) and no friction, determine
the velocity of each ball
immediately after impact.
Sample Problem 13.17
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A 30 kg block is dropped
from a height of 2 m onto
the 10 kg pan of a spring
scale. Assuming the impact
to be perfectly plastic,
determine the maximum
deflection of the pan. The
spring constant is k = 20
kN/m.